Electrical forces and energy: Electric Potential help

[Neutrinogun]

Homework Statement

Hi, I have a couple different problems that I did not understand.
I have the images here.
The problems which I did not understand were: 12, 13, 17. [EDIT: I calculated answer to 12 using conservation of energy.]
For 13, I know how the graph looks; however, I'm not sure how to calculate the two zeros as well as the local minimum on the left.

Homework Equations

The Attempt at a Solution

12. No idea whatsoever. I tried using the electrostatic forces equation, however there is no specific "r" value between where the Q2 and the other Q's are, so you can't do that. I think it might have something to do with momentum, but I'm not sure how to apply that.

EDIT: Uo + Ko = Uf + Kf. Using this, I got the right answer Vo = 6.19 m/s.

13. I already have the graph which looks like this:

The black spots are the values I need to find. I tried using the field graph to find the values, but it failed. I also tried using kq/r = kq/r to find the zeros, but I didn't get the right answers again.

17. This one I know how to do, I'm just confused on what to plug in.

The ΔV = -EΔx, so then you do Delta V (125 > 100) + Delta V (100 > 50) + Delta V (50>25) and that should be the answer. However I'm not sure what to plug in for the "q" in each case.

CORRECT ANSWERS
12. 6.19 m/s
13. The values of the two zeros are -5.46 m and 1.0 m. The y-value of the local min is -2.00 m.
17. 3.42 * 10^4 V

I'd appreciate it for any help. Thank you!

 

[anathema]

Hi there, let me see if I can help you with these problems.

12. For this problem, you are correct in using conservation of energy. The initial energy (Uo + Ko) is equal to the final energy (Uf + Kf). So, we can write the equation as Uo + Ko = Uf + Kf.

For the initial energy, we have the potential energy between Q1 and Q2, which is kQ1Q2/r. For the final energy, we have the kinetic energy, which is 1/2mv^2. So, the equation becomes kQ1Q2/r = 1/2mv^2.

Now, we can plug in the values given in the problem. We know that Q1 = 2q, Q2 = 4q, and r = 2m. We also know that the mass is 0.5 kg. So, we have:

k(2q)(4q)/2m = 1/2(0.5)(v^2)

Simplifying, we get:

8kq^2/m = 0.25v^2

Solving for v, we get:

v = √(8kq^2/m)

Plugging in the values for k, q, and m, we get v = 6.19 m/s.

13. For this problem, we need to find the zeros and the local minimum of the graph. To find the zeros, we need to set the y-value (or potential energy) equal to zero, since the graph crosses the x-axis at these points. So, we have:

kQ1Q2/r = 0

Solving for r, we get:

r = kQ1Q2/0

Since we cannot divide by zero, this means that the zeros are located at infinity. However, we can still find the values for the zeros using the given values in the problem. We know that Q1 = 4q and Q2 = 2q, so we have:

r = k(4q)(2q)/0

Simplifying, we get:

r = ∞

This means that the zeros are located at infinity, which is not a practical answer. So, we can use the given values to find the approximate values for the zeros. We can also use the graph to estimate the values. For the local